Question

Let $P\left(a_1, b_1\right)$ and $Q\left(a_2, b_2\right)$ be two distinct points on a circle with center $C(\sqrt{2}, \sqrt{3})$ Let $O$ be the origin and $OC$ be perpendicular to both $CP$ and $CQ$. If the area of the triangle $OCP$ is $\frac{\sqrt{35}}{2}$, then $a_1^2+a_2^2+b_1^2+b_2^2$ is equal to ______

Answer 1

$\frac{1}{2}​×PC×\sqrt5​=\frac{\sqrt{35}}{2}​​;PC=\sqrt7​$

$a^{2}_1​+b_{1}^2​+a^{2}_2​+b^{2}_2​=OP^2+OQ^2$
=2(5+7)=24
So , the correct answer is 24.